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I already did it but its a different problem type that I really was unsure about so I just want to verify my solution.
Problem: A 350-gram piece of metal at 100 C is dropped into a 100-gram aluminum cup containing 500 g of water at 15 C. The final temperature of the system is 40 C. What is the specific heat of the metal, assuming no heat is exchanged with the surroundings? Specific heat for aluminum is around 900 J/kg*K.
Basically:
m, m = .35kg
m, w = .5kg
m, a = .1kg
Ti, m = 100 C
Ti, a,w = 15 C
Tf = 40 C
Cp,w = 4186 J/kg*K
Cp,a = 900 J/kg*K
Cp,m = ?
Q,a = -Q,b
delta Q = m*Cp*delta T
how i did it:
Q,w,a = -Q,m
((4186 J/kg*K)(40-15)(.5) + (.1)(40-15)(900)) = -(.350)(C,m)(40-100)
52250 J + 2250 J = 21C,m
54500 = 21C,m
C,m = 2595.238 J/kg*K
Alright, step one here is really what I'm unsure about. I treated the aluminum and water as one part of the system, therefore I added the heat gained by the water and the aluminum as shown above, which I wasn't entirely sure of (in terms of that being allowed). Also I know a change in absolute temperature in K is the same as a change in temperature, deg C, but I was shaky on the interchangable use of kelvins and celsius degrees in the specific heats. I'm pretty sure that's valid, but I just would like to have that reaffirmed.
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Anyways, that corresponded pretty closely with one of the choices, 2600 J/kg*deg K so I'm just making sure the work is right because usually I'm only working with two objects in the system, not three. Thanks in advance.
------
Also: Can conductive heat transfer only occur when a solid can mediate the energy transfer? I think this is true, but I'm not sure. I'm loose about the methods of heat transfer, very loose.
Homework Statement
Problem: A 350-gram piece of metal at 100 C is dropped into a 100-gram aluminum cup containing 500 g of water at 15 C. The final temperature of the system is 40 C. What is the specific heat of the metal, assuming no heat is exchanged with the surroundings? Specific heat for aluminum is around 900 J/kg*K.
Basically:
m, m = .35kg
m, w = .5kg
m, a = .1kg
Ti, m = 100 C
Ti, a,w = 15 C
Tf = 40 C
Cp,w = 4186 J/kg*K
Cp,a = 900 J/kg*K
Cp,m = ?
Homework Equations
Q,a = -Q,b
delta Q = m*Cp*delta T
The Attempt at a Solution
how i did it:
Q,w,a = -Q,m
((4186 J/kg*K)(40-15)(.5) + (.1)(40-15)(900)) = -(.350)(C,m)(40-100)
52250 J + 2250 J = 21C,m
54500 = 21C,m
C,m = 2595.238 J/kg*K
Alright, step one here is really what I'm unsure about. I treated the aluminum and water as one part of the system, therefore I added the heat gained by the water and the aluminum as shown above, which I wasn't entirely sure of (in terms of that being allowed). Also I know a change in absolute temperature in K is the same as a change in temperature, deg C, but I was shaky on the interchangable use of kelvins and celsius degrees in the specific heats. I'm pretty sure that's valid, but I just would like to have that reaffirmed.
------
Anyways, that corresponded pretty closely with one of the choices, 2600 J/kg*deg K so I'm just making sure the work is right because usually I'm only working with two objects in the system, not three. Thanks in advance.
------
Also: Can conductive heat transfer only occur when a solid can mediate the energy transfer? I think this is true, but I'm not sure. I'm loose about the methods of heat transfer, very loose.
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